<< Chapter < Page Chapter >> Page >

You probably first encountered complex numbers when you studied values of z (called roots or zeros) for which the following equation is satisfied:

a z 2 + b z + c = 0 .

For a 0 (as we will assume), this equation may be written as

z 2 + b a z + c a = 0 .

Let's denote the second-degree polynomial on the left-hand side of this equation by p ( z ) :

p ( z ) = z 2 + b a z + c a .

This is called a monic polynomial because the coefficient of the highest-power term ( z 2 ) is 1. When looking for solutions to the quadratic equation z 2 + b a z + c a = 0 , we are really looking for roots (or zeros) of the polynomial p ( z ) . The fundamental theorem of algebra says that there are two such roots. When wehave found them, we may factor the polynomial p ( z ) as follows:

p ( z ) = z 2 + b a z + c a = ( z - z 1 ) ( z - z 2 ) .

In this equation, z 1 and z 2 are the roots we seek. The factored form p ( z ) = ( z - z 1 ) ( z - z 2 ) shows clearly that p ( z 1 ) = p ( z 2 ) = 0 , meaning that the quadratic equation p ( z ) = 0 is solved for z = z 1 and z = z 2 . In the process of factoring the polynomial p ( z ) , we solve the quadratic equation and vice versa.

By equating the coefficients of z 2 , z 1 , and z 0 on the left-and right-hand sides of [link] , we find that the sum and the product of the roots z 1 and z 2 obey the equations

z 1 + z 2 = - b a z 1 z 2 = c a .

You should always check your solutions with these equations.

Completing the Square. In order to solve the quadratic equation z 2 + b a z + c a = 0 (or, equivalently, to find the roots of the polynomial z 2 + b a z + c a ) , we “complete the square” on the left-hand side of [link] :

( z + b 2 a ) 2 - ( b 2 a ) 2 + c a = 0 .

This equation may be rewritten as

( z + b 2 a ) 2 = ( 1 2 a ) 2 ( b 2 - 4 a c ) .

We may take the square root of each side to find the solutions

z 1 , 2 = - b 2 a ± 1 2 a b 2 - 4 a c .

In the equation that defines the roots z 1 and z 2 , the term b 2 - 4 a c iscritical because it determines the nature of the solutions for z 1 and z 2 . In fact, we may define three classes of solutions depending on b 2 - 4 a c .

(i) Overdamped ( b 2 - 4 a c > 0 ) . In this case, the roots z 1 and z 2 are

z 1 , 2 = - b 2 a ± 1 2 a b 2 - 4 a c .

These two roots are real, and they are located symmetrically about the point - b 2 a . When b = 0 , they are located symmetrically about 0 at the points ± 1 2 a - 4 a c . (In this case, - 4 a c > 0 . ) Typical solutions are illustrated in [link] .

There are three cartesian graphs in this series of images. The first graph has two points on the negative portion of the x axis labeled z_2 and z_1 proceeding to the the left. In between these two points there is another point labeled -b/2a. The second graph is similar except that the point z_2 is located on the positive side of the x-axis while the point z_1 is still present at the far extreme of the negative portion of the y axis. The point -b/2a is located between these points but this time is closer to the origin. The third graph does not contain the point -b/2a, but z_1 is located on the far extreme of the negative portion of the x axis and the point z_2 is located on the far extreme of the positive x axis. There are three cartesian graphs in this series of images. The first graph has two points on the negative portion of the x axis labeled z_2 and z_1 proceeding to the the left. In between these two points there is another point labeled -b/2a. The second graph is similar except that the point z_2 is located on the positive side of the x-axis while the point z_1 is still present at the far extreme of the negative portion of the y axis. The point -b/2a is located between these points but this time is closer to the origin. The third graph does not contain the point -b/2a, but z_1 is located on the far extreme of the negative portion of the x axis and the point z_2 is located on the far extreme of the positive x axis.
Typical Roots in the Overdamped Case; (a) b / 2 a > 0 , 4 a c > 0 , (b) b / 2 a > 0 , 4 a c < 0 , and (c) b / 2 a = 0 , 4 a c < 0

(ii) Critically Damped ( b 2 - 4 a c = 0 ) . In this case, the roots z 1 and z 2 are equal (we say they are repeated):

z 1 = z 2 = - b 2 a .

These solutions are illustrated in [link] .

This Cartesian graph contains two points that rather close together. Both points exist on the negative portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the negative portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the positive portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the positive portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a.
Roots in the Critically Damped Case; (a) b / 2 a > 0 , and (b) b / 2 a < 0

(iii) Underdamped ( b 2 - 4 a c < 0 ) . The underdamped case is, by far, the most fascinating case. When b 2 - 4 a c < 0 , then the square root in the solutions for z 1 and z 2 ( [link] ) produces an imaginary number. We may write b 2 - 4 a c as - ( 4 a c - b 2 ) and write z 1 , 2 as

z 1 , 2 = - b 2 a ± 1 2 a - ( 4 a c - b 2 ) = - b 2 a ± j 1 2 a 4 a c - b 2 .

These complex roots are illustrated in [link] . Note that the roots are

purely imaginary when b = 0 , producing the result

z 1 , 2 = ± j c a .
This Cartesian graph contains a line segment that extends from the origin up and to the left into quadrant II. The ending point is labeled x and the line appears to be labeled r. An arch originates on the positive portion of the x axis and ends near the the middle of the line that was just mentioned. This arch is labeled θ. There is a point on the upper end of the positive portion of the y-axis labeled 1/2a sqrt(4ac-b^2). There is also a point in the middle of the negative portion of the x axis labeled -b/2a. Directly below this point there is a marked by an x. This Cartesian graph contains a line segment that extends from the origin up and to the left into quadrant II. The ending point is labeled x and the line appears to be labeled r. An arch originates on the positive portion of the x axis and ends near the the middle of the line that was just mentioned. This arch is labeled θ. There is a point on the upper end of the positive portion of the y-axis labeled 1/2a sqrt(4ac-b^2). There is also a point in the middle of the negative portion of the x axis labeled -b/2a. Directly below this point there is a marked by an x.  This Cartesian graph contains has a point  just above the origin on the y axis and it is labeled 1/2a sqrt(4ac-b^2). On the positive portion of the x axis there is a point. Above and below this point there is an x. At the very end of the x axis is the fraction -b/2a.  This Cartesian graph contains has a point  just above the origin on the y axis and it is labeled 1/2a sqrt(4ac-b^2). On the positive portion of the x axis there is a point. Above and below this point there is an x. At the very end of the x axis is the fraction -b/2a. This Cartesian graph has a point on the positive and negative portion of the y axis. Both points are the same distance from the origin. The upper point is labeled sqrt(c/a). This Cartesian graph has a point on the positive and negative portion of the y axis. Both points are the same distance from the origin. The upper point is labeled sqrt(c/a).
Figure 1.12: Typical Roots in the Underdamped Case; (a) b / 2 a > 0 , ( b ) b / 2 a < 0 , and (c) b / 2 a = 0

In this underdamped case, the roots z 1 and z 2 are complex conjugates:

z 2 = z 1 * .

Thus the polynomial p ( z ) = z 2 + b a z + c a = ( z - z 1 ) ( z - z 2 ) also takes the form

p ( z ) = ( z - z 1 ) ( z - z 1 * ) = z 2 - 2 Re [ z 1 ] z + | z 1 | 2 .

Re [ z 1 ] and | z 1 | 2 are related to the original coefficients of the polynomial as follows:

2 Re [ z 1 ] = - b a
| z 1 | 2 = c a

Always check these equations.

Let's explore these connections further by using the polar representations for z 1 and z 2 :

z 1 , 2 = r e ± j θ .

Then [link] for the polynomial p ( z ) may be written in the “standard form”

p ( z ) = ( z - r e j θ ) ( z - r e - j θ ) = z 2 - 2 r c o s θ z + r 2 .

[link] is now

2 r cos θ = - b a r 2 = c a

These equations may be used to locate z 1 , 2 = r e ± j θ

r = c a θ = ± cos - 1 ( - b 4 a c ) .

Questions & Answers

give me the waec 2019 questions
Aaron Reply
the polar co-ordinate of the point (-1, -1)
Sumit Reply
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
Rockstar Reply
tanh`(x-iy) =A+iB, find A and B
Pankaj Reply
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
Branded Reply
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
archana Reply
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
Lhorren Reply
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
jancy Reply
answer
Ajith
exponential series
Naveen
what is subgroup
Purshotam Reply
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
Vinay Reply
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
Payal Reply
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
Leonardo Reply
How do i figure this problem out.
how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
Difference between extinct and extici spicies
Amanpreet Reply
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
Mapo Reply
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
Mapo Reply
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, A first course in electrical and computer engineering. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10685/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'A first course in electrical and computer engineering' conversation and receive update notifications?

Ask
Anindyo Mukhopadhyay
Start Quiz